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Proof step 2 from proof sub-insert (previous step)

s ⊆ s0 ++ a ⊦ a ∈ s, s ⊆ s0, s0 ++ a ⊆ s0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
s ⊆ s0 ++ a ⊦ a ∈ s0, a ∈ s, s ⊆ s0, s0 ++ a ⊆ s0

Rule argument

fma: a ∈ s0

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%sub-idem: ⊦ s ⊆ s; Used formulas: a ∈ s0